Equation of Circle — Standard, general form
Circles
8
JEE Qs
8%
Hard
60
min
Master the interconversion between standard and general forms and their properties to quickly derive center, radius, and conditions for various circle types.
🧮 Key Formulas
✅ Key Points for JEE
- 1The general equation of a second-degree curve `Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0` represents a circle if the coefficients of x² and y² are equal (`A=B`) and non-zero, and the xy term is absent (`H=0`).
- 2The standard form `(x-h)^2 + (y-k)^2 = r^2` directly gives the center `(h, k)` and radius `r`, making it useful when these properties are known or easily derived.
- 3The general form `x^2 + y^2 + 2gx + 2fy + c = 0` is particularly useful when three points on the circle are given, as these can be substituted to form a system of linear equations in g, f, and c.
- 4For a real circle, the radius `r` must be positive, implying `g^2 + f^2 - c > 0`. If `g^2 + f^2 - c = 0`, it is a point circle; if `g^2 + f^2 - c < 0`, it is an imaginary circle.
- 5The diameter form `(x-x1)(x-x2) + (y-y1)(y-y2) = 0` is a quick way to find the equation of a circle when the coordinates of the endpoints of a diameter `(x1, y1)` and `(x2, y2)` are given, leveraging the property that the angle in a semicircle is 90 degrees.
⚠️ Common Mistakes
- ✕Failing to normalize the general equation by dividing by the coefficient of `x^2` and `y^2` (if not 1) before extracting `g`, `f`, and `c` values.
- ✕Sign errors when determining the center `(-g, -f)` from the general form, or when substituting `c` into the radius formula `sqrt(g^2 + f^2 - c)`.
- ✕Incorrectly identifying `2g` and `2f` terms if the `x` or `y` term coefficients are not explicitly even (e.g., `5x` implies `2g=5`, so `g=5/2`).
📝 Practice Questions
See allQ1. Let circle C be the image of x2 + y2 −2x + 4y −4 = 0 in the line 2x −3y + 5 = 0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C . If B(α, β), with β < 4, lies on C such that the length of the are AB is (1/6)th of the perimeter of C , then β −√3α is equal to (1) 3 + √3 (2) 4 (3) 4 −√3 (4) 3
Q19.Let the line x + y = 1 meet the circle x2 + y2 = 4 at the points A and B . If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D , then the area of the quadrilateral ADBC is equal to : (1) √14 (2) 3√7 (3) 2√14 (4) 5√7
Q18.A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2, 5) and intersects the circle C at exactly two points. If the set of all possible values of r is the interval (α, β), then 3β −2α is equal to : (1) 10 (2) 15 (3) 12 (4) 14
Q21.Let the circle C touch the line x −y + 1 = 0, have the centre on the positive x -axis, and cut off a chord of length 4 along the line −3x + 2y = 1. Let H be the hyperbola x2 −y2 = 1, whose one of the foci is the √13 α2 β2 centre of C and the length of the transverse axis is the diameter of C . Then 2α2 + 3β2 is equal to ______
Q6. Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0 and cuts off an intercept of length b on y-axis be x2 + y2 −αx + βy + γ = 0. If the circle lies below x-axis, then the ordered pair (2a, b2) is equal to (1) (γ, β2 −4α) (2) (α, β2 + 4γ) (3) (γ, β2 + 4α) (4) (α, β2 −4γ) 2x
Q15.Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C , whose mid-point is (1, 2), is : (1) √3 (2) 2√2 (3) 2√3 (4) 4√2
NCERT Chapters
- Class 11 Maths Ch 11: Conic Sections